# Binomial coefficient help?

I'm studying for my exams and would appreciate any help with binomial coefficients. I think I got the idea but having trouble with a specific one:

Q) If a there are 11 dogs and 9 cats:

a) How many 7 different groups can be made?

b) How many 8 different groups can made that have at least 2 cats?

c) How many 9 different groups can be made that have at least 2 cats and 2 dogs?

For this one I have no idea, I'm guessing $\binom{20}{8}$-($\binom{9}{0}$*$\binom{11}{8}$)-($\binom{9}{1}$*$\binom{11}{7}$)-($\binom{11}{0}$*$\binom{9}{9}$)-($\binom{11}{1}$*$\binom{9}{8}$). If it's not, mind explaining how to do it? If it is, mind explaining why this works? I completely guessed here.

d) How many 6 different groups can be made that have exactly 4 cats?

-
You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. In particular, type $\binom{n}{k}$ to get $\binom{n}{k}$ –  Zev Chonoles Apr 8 '13 at 4:49
Oh cool, never knew that, thanks :)! Hope it looks better now. –  StackPWRequirmentsAreCrazy Apr 8 '13 at 4:56
The expressions for a), b), d) are correct. (I have not done the numerical calculations.) For c) the idea is right but there are typos. Of course it should be $\binom{20}{9}$ in front, you know that. And for the correction terms, bottoms should each add up to $9$. So the last two correction terms are right, the first two are not. –  André Nicolas Apr 8 '13 at 5:01
Thank you very much Andre for your help :). –  StackPWRequirmentsAreCrazy Apr 8 '13 at 5:13

## 1 Answer

You did fine with (a), (b), and (d). Your approach to (c) is also fine, but you kept switching back and forth between groups of $8$ and groups of $9$ animals. There are $\binom{11}9$ ways to choose a group with no cats, and $\binom{11}8\binom91$ ways to choose a group with exactly one cat. There is only one way to choose a group of $9$ with no dogs, and there are $\binom{11}1\binom98$ ways to choose a group with exactly one dog. That’s a grand total of

$$\binom{11}9+9\binom{11}8+1+11\cdot9=55+1485+1+99=1640$$

‘bad’ groups out of $\binom{20}9=167,960$ possible groups, for a total of $166,320$ acceptable groups.

-
Thank you very much for your reply :). Mind explaining where the 1 came from? I think it's a simple step that I'm not seeing. –  StackPWRequirmentsAreCrazy Apr 8 '13 at 5:05
@StackPWRequirmentsAreCrazy: You’re very welcome. You had it yourself: $\binom{11}0\binom99=1$. I just avoided the binomial coefficients by observing that since there are only $9$ cats, a group of $9$ with no dogs must simply be the group consisting of all $9$ cats. –  Brian M. Scott Apr 8 '13 at 5:09
Oh, I see, my bad. Thanks once again for all your help! –  StackPWRequirmentsAreCrazy Apr 8 '13 at 5:12
@StackPWRequirmentsAreCrazy: My pleasure! –  Brian M. Scott Apr 8 '13 at 5:13